﻿ Class 9 NCERT Math Solution
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Exercise - 1.5

Question-1 :-  Classify the following numbers as rational or irrational:
(i) 2 - √5, (ii) (3+√23) - √23, (iii) 2√7/7√7, (iv) 1/√2, (v) 2π

Solution :-
```(i)   2 - √5 = 2 - 2.236067... = 0.236067
As the decimal expansion of this expression is non-terminating non-recurring,
therefore, it is an irrational number.

(ii)  (3+√23) - √23 = 3+√23-√23 = 3 =3/1
therefore, it is a rational no. and it can be represented in p/q.

(iii) 2√7/7√7 = 2/7
therefore, it is a rational no. and it can be represented in p/q.

(iv)  1/√2 = 1/√2 x √2/√2 = √2/2 = 0.7071067811...
As the decimal expansion of this expression is non-terminating non-recurring,
therefore, it is an irrational number.

(v)  2π = 2 x 3.1415... = 6.2830...
As the decimal expansion of this expression is non-terminating non-recurring,
therefore, it is an irrational number.
```

Question-2 :- Simplify each of the following expressions:
(i) (3+√3)(2+√2), (ii) (3+√3)(3-√3), (iii) (√5+√2)², (iv) (√5-√2)(√5+√2)

Solution :-
```(i)   (3+√3)(2+√2)
=3(2+√2)+√3(2+√2)
=6+3√2+2√3+√6

(ii)  (3+√3)(3-√3)
=3²-(√3)²
=9-3 = 6

(iii) (√5+√2)²
=(√5)²+(√2)²+2√5√2
=5+2+2√10
=7+2√10

(iv)  (√5-√2)(√5+√2)
=(√5)²-(√2)²
=5-2 = 3
```

Question-3 :-  Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Solution :-
``` There is no contradiction. When we measure a length with scale or any other instrument,
we only obtain an approximate rational value.We never obtain an exact value.
For this reason, we may not realise that either c or d is an irrational.
Therefore, the c/d fraction is irrational. Hence, π is an irrational.
```

Question-4 :- Represent √9.3 . on the number line.

Solution :-
``` Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit.
Find the midpoint D of OC and draw a semi-circle on OC while taking D as its centre.
Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E.
Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is √9.3 ```

Question-5 :- Rationalise the denominators of the following:
(i) 1/√7, (ii) 1/(√7-√6), (iii) 1/(√5+√2), (iv) 1/(√7-2)

Solution :-
```(i)   1/√7   Here 1 is the Numenator and √7 is Denominator, so multiply and divide by √7
=1/√7 x √7/√7
=√7/7

(ii) 1/(√7-√6)   Here 1 is the Numenator and (√7-√6) is Denominator, so multiply and divide by (√7+√6)
=1/(√7-√6) x (√7+√6)/(√7+√6)
=(√7+√6)/(7-6)
=(√7+√6)/1 = √7+√6

(iii) 1/(√5+√2)  Here 1 is the Numenator and (√5-√2) is Denominator, so multiply and divide by (√5-√2)
=1/(√5+√6) x (√5-√2)/(√5-√2)
=(√5-√2)/(5-2)
=(√5-√2)/3

(iv) 1/(√7-2)   Here 1 is the Numenator and (√7-2) is Denominator, so multiply and divide by (√7+2)
=1/(√7-2) x (√7+2)/(√7+2)
=(√7+2)/(7-4)
=(√7+2)/3
```
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